Question

# A large punch bowl holds 3.70 kg of lemonade (which is essentially water) at 25.0 ∘C....

A large punch bowl holds 3.70 kg of lemonade (which is essentially water) at 25.0 ∘C. A 6.00×10−2-kg ice cube at -11.0 ∘C is placed in the lemonade.

What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings.

What is the amount of ice (if any) remaining?

here

Specific heat of water = 4181.3 J/(kg.K)
Specific heat of ice = 2110 J/(kg.K)
Latent heat of fusion for water = 334000 J/kg

Energy absorbed by ice warming to 0 deg C
= 2110 * 0.06 * 11 = 1392.6 J

Energy released by lemonade cooling to 0 deg C
= 4181.3 * 3.7 * 25 = 386770.25 J

Surplus energy from lemonade in warming the ice to 0 deg C
= 386770.25 - 1392.6 = 385377.65 J

This energy can melt M kg of ice
M * 334000 = 385377.65
M = 385377.65/334000 = 1.153 kg

Ice remaining unmelted
= 0.06 - 1.153 = -1.093 kg
whole ice will be melted . no ice will be remaining

and the final temperature is

3.7 * T = 0.06 * (25 - T)

T = 0.3989 C the ice melts to this temperature